CAT 2008 Gurgaon Study Group

[Dream08]

Originally Posted by catguru999

well i found the class really good...
and those theorems fermat, chinese and others :i think they are important..
i too went home and saw the CAT 2007 paper:
1. the question related to "confused bankteller" where he interchanges the denomination was nothing but application of "chinese theorem".
2. there was a question in CAT 07: a set s is given {2,3,4, ..., 2n +1} and ..... < rest you can see from the paper>
i think in this question his trick works.. i dont think there was anything to do with 2007 here.. so just put n = 1 and then n = 2 and you will get the answer.
3. also another question with function f(n) was there where you really needed to do speed calculations...
i am sure there will be more of them...
anyways i am going for the next class on saturday morning.. i will advise you to come and give it a try...

Hello Guru,
I think I will join Byju anyway. I have not been taking any coaching for CAT; just taking the mock cats. Also, it does not hurt learning few tricks though.
Regarding the questions that you mentioned, I looked into the CAT2007 paper again but could not find any link between the questions and the method that Byju taught in the class.
i) The confused banker question is not a chinease thm application. It's just this equation 100b + a -50 = 3(100a + b)
where the original amount was something like a rupees b paise i.e. 100a + b paise. if you take a to be 4 then b comes out to be less than 100 paise i.e 1 Rs. and hence the statement 5 is correct.
ii) Yes you are correct, it does not matter what the value of n could be. Yes I solved this question on the D day by taking n = 2008 . Solved it correctly but took a lot of time, 90 sec approx. The thought that n does not matter here did not even occur to me. Now I know, you can always ignore n as long as same function is being used for all values of n.

Let's meet on Saturday at Byju's classes. By the way it was a good experience attending classes after such a long, very long time. Nice to see some new faces there. I guess you know what I mean .

CAT2008(ABC). Here I come at Jet speed!!

[catguru999]

Originally Posted by Sandy2003

Hello Friends,

Can anyone of u tell me where are these classes held, what is the fees for full course???

these classes are being held somewhere in Gurgaon and the fees is Rs.9000/-
for details you can send me a private message...

[masoom]

hi puys...

i am working in TCS gurgaon ( 1year exp) ....

i have joined TIME weekend course and CL Test series...

I want to join in this Study Group

[siddharthpareek]

Originally Posted by masoom

hi puys...

i am working in TCS gurgaon ( 1year exp) ....

i have joined TIME weekend course and CL Test series...

I want to join in this Study Group

Hi Masoom,
welcome to gurgaon study group..well do see this thread for the updates going on...we might plan for a meet up this weekend i.e on Saturday or Sunday..i will soon post the details for the same..if you have any questions you can PM me.

[hi.shivani]

Some one solve these problems:
Q1) For which of the values of g(x) can f(x) be expressed as f1(x) + f2(x) where f1(x) & f2(x) are any two function of x?

a) g (x) = ex b) g (x)=x2 c) g (x) =log(x) d) None of these

Q2) If 0<x<1000 & [x/2]+[x/3]+[x/5]=31x/30, where[x] is the greatest integer less than or equal to x,the number of possible values of x is,
a) 34 b) 32 c) 33 d) None of these

Q3) (x2-1) is a factor of f(x)=(x5+ax4+bx3+dx2+x+d).The graph of f(x) intersects Y axis at (0,-3). Find the value of (a+c).
a) 0 b) 3 c) -3 d) -1

Q4) Let f(x) =ax2+bx+c,where a,b,c are rational & f:z->,where z is the set of integers,which of the following best describes the value of a+b

a) a negative integer
b) an integer
c) not integral ratonal number
d) None of these

Q5) If f(x+1)+f(x-1)=2f(x) & f(0) =0,then f(x),n is a natural number is
a) nf(1) b) {f(1)}^n c) 0 d) None of these

Q6) The domainof the function f(x) =16-xC2x-1 + 20-3xP4x-5,
is the set
a) [1,2,3,4,5] b) {2,3,4} c) {2,3} d) None of these

Q7) The domain of the function f(x)=loge[x-mod(x)] where [.] denotes the greatest integeral function is
a) R b) R-Z c) (0,infinity) d) None of these

QIf f(x) =x^n, n is a natural number g of (x) =n g(x),then g(x) can be
a) n[mod(x)] b) 3(cube rootx) d) e^x d) log [mod(x)]

Q9)If f(x) =4x/(4x+2),find the value of f(1/1999)+f(2/1999)+...+f(1998/1999)
a) 1998 b) 1999 c) 998 d)999

Q10) If f(x)=2x2+6x-1,then the value of [f(3/4)+1]/[f(3/4)-1]
a) 11/13 b) 35/3 c) 45/29 d) None of these

Q11) Find f(111)
a) 1 b) -1 c) 2 d) None of these

Q12) f(1)+f(2)+...+f(25)
a) -26 b) -24 c) -22 d) None of these

Q13) Let y=f(x)=loga x and a>1. Then only one of the solution is
a) If x=1 then y=0
b) If x<1 then y<0
c) If x=1/2 then y=1/2
d) If x=1 then y=1.

14) A function H is defined for all positive integers that satisfy the following condition
a) 2/2005 b)2/2006 c) 2/2006! d)2/2005!

[masoom]

@ shivani

1. we know that log(a*b)= log (a) +log(b)

hence if g(x)=log(x)
=> g(x)= 1/2log(x) + 1/2log(x)
g(x)=f1(x) +f2(x).

3. f(x)=x5+ax4+bx3+cx2+x+d

now graph passes through(0,-3)
thus f(0)=-3

hence d=-3

now x2-1 is a root of f(x)
thus f(1) and f(-1) will be zero..
and hence... by solving we get b=-2
and a+c=3

[masoom]

5. i am getting none of these as answer


8. gof(x)= g(f(x)

means in exprexxion of g(x) substitute x by f(x)

if we do this in each option provided

then g(x) = log[mod(x) will satisfy


13. b

property of log is

y=log(a)b where a is base

then y>0 for b>a and vice versa

[masoom]

is the answer for 9th is 998...

[masoom]

ans 2:

[x/2]+[x/3]+[x/5]=31x/20

now we can write [x]= x-{x}

where {x} is the fractional part of x.

using this property in the given equation

x/2 + x/3 + x/5 - {x/2}-{x/3}-{x/5}=31x/30

we get , {x/2}+{x/3}+{x/5}=0

since fractional part of each term is zero thus x should be the multiple of LCM of 2.3 and 5 eg 30.

so possible valuew are 30,60,90,.....,990

total 33 values uptill 1000.

[hi.shivani]

@ masoom

ans to:
1) None of these
5) a) nf(1)

rest wat u hv soved z coorect but i hv 2 confusions

1) in q 3 plz do elaborate hw did u got f(0)=-3.
See its given f(x)=ax^4+bx^3+cx+d
f(0)=0+0+0+d
=d

And in question 2) explain this portion i m nt understanding"since fractional part of each term is zero thus x should be the multiple of LCM of 2.3 and 5 eg 30."